How Do You Calculate Final Velocities in a Two-Dimensional Collision?

[alisa7217]

I need help. I don't know how to do this problem becuase i need to solve for both velicty final of a and b and i keep screwing it up, any help you can give me i would appreciate.

"[CJ5 7.P.032.] The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.029 kg and is moving along the x-axis with a velocity of +5.5 m/s. It makes a collision with puck B, which has a mass of 0.049 kg and is initially at rest. After the collision, the two pucks fly apart. A flys off the x-axis with a 65 degree angle and B flys off the x-axis with a 37 degree angle in the negative y direction.


Find the final speed of
(a) puck A and

(b) puck B.

 

[OlderDan]

I need help. I don't know how to do this problem becuase i need to solve for both velicty final of a and b and i keep screwing it up, any help you can give me i would appreciate.

"[CJ5 7.P.032.] The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.029 kg and is moving along the x-axis with a velocity of +5.5 m/s. It makes a collision with puck B, which has a mass of 0.049 kg and is initially at rest. After the collision, the two pucks fly apart. A flys off the x-axis with a 65 degree angle and B flys off the x-axis with a 37 degree angle in the negative y direction.


Find the final speed of
(a) puck A and

(b) puck B.

You will assume no loss of energy (an elastic collision) Before the collision puck 1 has momentum in its direction of motion, and all the energy (kinetic). After the collision the momentum of the two pucks must equal the original momentum of the first puck and the sums of their energies must equal the initial energy of puck 1. Momentum is a vector. You must break momentum of each puck into two components, one in the direction of the original motion and one perpendicular to that direction.

 

[thepatientmental]

First, it is important to understand the concept of conservation of momentum in two-dimensional collisions. This means that the total momentum of the system before the collision is equal to the total momentum after the collision. The total momentum is calculated by multiplying the mass of an object by its velocity.

For this problem, we can use the equations for conservation of momentum in the x and y directions separately. In the x-direction, we have:

(mass of A)(initial velocity of A) = (mass of A)(final velocity of A) + (mass of B)(final velocity of B)

Plugging in the values given in the problem, we get:

(0.029 kg)(5.5 m/s) = (0.029 kg)(final velocity of A) + (0.049 kg)(final velocity of B)

Solving for the final velocity of A, we get:

final velocity of A = (0.029 kg)(5.5 m/s) - (0.049 kg)(final velocity of B) / 0.029 kg

In the y-direction, we have:

(mass of A)(initial velocity of A) = (mass of A)(final velocity of A) + (mass of B)(final velocity of B)

Plugging in the values given in the problem, we get:

(0.029 kg)(0 m/s) = (0.029 kg)(final velocity of A) + (0.049 kg)(final velocity of B)

Solving for the final velocity of B, we get:

final velocity of B = (0.029 kg)(final velocity of A) + (0.049 kg)(final velocity of B) / 0.049 kg

Now, we can use the given angles to find the x and y components of the final velocities. In the x-direction, the velocity of A will be:

velocity of A in the x-direction = (final velocity of A)(cos 65 degrees)

And in the y-direction, the velocity of B will be:

velocity of B in the y-direction = (final velocity of B)(sin 37 degrees)

Finally, we can use the Pythagorean theorem to find the magnitude of the final velocities:

final speed of A = √(velocity of A in x-direction)^2 + (velocity of A in y-direction)^2

final speed of B = √(velocity of B in x-direction)^2 + (velocity of B in